Simply supported beam, partial uniform load

[budgie69]

I was looking on the net using the search words listed in the subject line.
This eng tips forum came up, so I had a look.

MrMikee posted 23 June 06 16:29 with a formula from Blodgett "design of welded structures".

From Blodgett "Design of Welded Structures" Case 3c for beam supported at both ends with uniform load partially distributed over span:

when a=c the deflection at center is... wb(8L3-4b2L+b3)/(384EI)

When a=c uniform load is length 'b' centered on span.

w = unit load
L = length
E and I as you would expect

I don't think this works. If you go through the units you are left with mm2, not mm. Can anyone with access to Blodgett confirm this is the correct formula? It would be handy to have a formula for this situation.
Cheers

 

[rb1957]

are you sure w is a point load or is distributed ? ... if w is lb/in then you're ok with units, yes?

another day in paradise, or is paradise one day closer ?

 

[chicopee]

I don't have that book but I tend to agree with rb1957. Lower case "w" normally implies distributed load at least in all my reference books.

 

[budgie69]

Thanks for the feedback.

I checked thru the units again and yes if i use W = n/mm then units cancel out to mm. I had checked this but obviously, not very well. Makes it simple now if you have the load, just use Newtons and don't multiply by b.
Only need to multiply by B if your load is N/mm.

Thanks for taking the trouble to respond.

 

[rb1957]

W = wb

another day in paradise, or is paradise one day closer ?